Question

$ABCD$ is a trapezium in which $AB\parallel DC,BD$ is a diagonal and $E$ is the mid-point of them. A line is drawn through $E$ parallel to $AB$ intersecting $BC$ and $F$. Then $F$ is the mid-point of $BC$.

• A. True
• B. False

Answer 1

The correct option is A True

In $\Delta DAB$ and $\Delta DEG$

$\angle D$ is common

$\angle DBA=\angle DGE$ [corresponding angles $EF\parallel AB$]

Thus, $\Delta DAB\simeq \Delta DEG$

$\therefore \frac{DE}{DA}=\frac{DB}{DG}$

$\Rightarrow \frac{DB}{DG}=\frac{1}{2}$

Thus G is the mid point of BD.

Similarly in $\Delta BCD$ and $\Delta BFG$

$\angle B$ is common

$\angle BCD=\angle BFG$ [corresponding angles $EF\parallel CD$]

Thus, $\Delta BCD\simeq \Delta BFG$

$\therefore \frac{BD}{BG}=\frac{BC}{BF}$

$\Rightarrow \frac{BD}{BF}=\frac{1}{2}$

Thus F is the mid point of BC.